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\(D=\dfrac{5}{1\cdot2}+...+\dfrac{5}{199\cdot200}\)
\(=\dfrac{5}{2}\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{199}-\dfrac{1}{200}\right)\)
\(=\dfrac{5}{2}\cdot\dfrac{199}{200}=\dfrac{199}{80}\)
Lời giải:
\(D=5\times \left(\frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+...+\frac{1}{199\times 200}\right)\)
\(=5\times \left(\frac{2-1}{1\times 2}+\frac{3-2}{2\times 3}+\frac{4-3}{3\times 4}+...+\frac{200-199}{199\times 200}\right)\)
\(=5\times \left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{199}-\frac{1}{200}\right)=5\times (1-\frac{1}{200})\)
\(=5\times \frac{199}{200}=\frac{995}{200}=\frac{199}{40}\)
\(E=\dfrac{0.5}{1.2}+\dfrac{0.5}{2\cdot3}+...+\dfrac{0.5}{199\cdot200}\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{200}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{199}{200}=\dfrac{199}{400}\)
`x+0,25=18/5+43/4`
`x+0,25=3,6+10,75`
`x=3,6+10,5`
`x=14,1`
Vậy `x=14,1`
`x+0,25=18/5+43/4`
`⇔x+1/4=18/5+43/4`
`⇔ x-18/5=43/4-1/4`
`⇔ x-18/5=42/4=21/2`
`⇔ x-36/10=105/10`
`⇔ x=141/10`
\(\frac{1}{2}:0,5-\frac{1}{4}:0,25+\frac{1}{8}:0,125-\frac{1}{10}:0,1\)
\(=\frac{1}{2}\times2-\frac{1}{4}\times4+\frac{1}{8}\times8-\frac{1}{10}\times10\)
\(=1-1+1-1\)
\(=0\)
`4 xx 0,25 xx 5 xx 1/5 xx 2 xx 1/2`
`=(4xx0,25) xx(5xx1/5) xx (2xx1/2)`
`= 1 xx 5/5 xx 2/2`
`= 1 xx 1 xx 1`
`=1`
`@ M``i``n`
1)\(y\times7:5+4\times8=134\)
\(\Leftrightarrow y\times7:5+32=134\)
\(\Leftrightarrow y\times7:5=102\)
\(\Leftrightarrow y\times7=510\)
\(\Leftrightarrow y=72,86\)
2) \(\dfrac{1}{4}:0,25-\dfrac{1}{8}:0,125+\dfrac{1}{2}:0,5-\dfrac{1}{10}\)
\(=0,25:0,25-0,125:0,125+0,5:0,5-\dfrac{1}{10}\)
\(=1-1+1-\dfrac{1}{10}\)
\(=\dfrac{9}{10}\)
\(H=0,25\times\left(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+...+\dfrac{1}{19\times20}\right)\)
\(=0,25\times\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{19}-\dfrac{1}{20}\right)\)
\(=0,25\times\left(1-\dfrac{1}{20}\right)=0,25\times\dfrac{19}{20}=\dfrac{19}{80}\)
\(H=\dfrac{0.25}{1\cdot2}+\dfrac{0.25}{2\cdot3}+...+\dfrac{0.25}{199\cdot200}\)
\(=\dfrac{1}{4}\cdot\dfrac{199}{200}=\dfrac{199}{800}\)